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SuicideFuel Math thread problem (official)

How many anagrams, whose consonants are not together, can be formed with the word EFICIALATE
 
How many anagrams, whose consonants are not together, can be formed with the word EFICIALATE
We distinguish four cases:
  • Neither the first nor the last letter is a consonant. Using stars and bars we find that there are 4!*6!/(2*2*2)*(6-1 choose 4) such anagrams (the 4! accounts for all the possible ways to order the 4 consonants and the 6!/(2*2*2) accounts for all the possible ways to order the 6 vowels (where we need to divide by 2 thrice because all three distinct vowels appear twice))
  • Only the first letter is a consonant. Using stars and bars again we find that there are 4!*6!/(2*2*2)*(6-1 choose 3) such anagrams
  • Only the last letter is a consonant. By symmetry there are 4!*6!/(2*2*2)*(6-1 choose 3) such anagrams as well
  • Both the first and the last letter are consonants. Using stars and bars one last time we find that there are 4!*6!/(2*2*2)*(6-1 choose 2) such anagrams
All in all, there are therefore 4!*6!/(2*2*2)*((5 choose 1)+3*(5 choose 2)) = 3*6!*(5+3*10) = 3*6!*35 = 15*7! = 15*5040 = 75600 anagrams without consecutive consonants
 
Last edited:
We distinguish four cases:
  • Neither the first nor the last letter is a consonant. Using stars and bars we find that there are 4!*6!/(2*2*2)*(6-1 choose 4) such anagrams (the 4! accounts for all the possible ways to order the 4 consonants and the 6!/(2*2*2) accounts for all the possible ways to order the 6 vowels (where we need to divide by 2 thrice because all three distinct vowels appear twice))
  • Only the first letter is a consonant. Using stars and bars again we find that there are 4!*6!/(2*2*2)*(6-1 choose 3) such anagrams
  • Only the last letter is a consonant. By symmetry there are 4!*6!/(2*2*2)*(6-1 choose 3) such anagrams as well
  • Both the first and the last letter are consonants. Using stars and bars one last time we find that there are 4!*6!/(2*2*2)*(6-1 choose 2) such anagrams
All in all, there are therefore 4!*6!/(2*2*2)*((5 choose 1)+3*(5 choose 2)) = 3*6!*(5+3*10) = 3*6!*35 = 15*7! = 15*5040 = 75600 anagrams without consecutive consonants
Correct :feelsokman:
 
y=|lnx|

Find the first order derivative of the above function
 
We distinguish four cases:
  • Neither the first nor the last letter is a consonant. Using stars and bars we find that there are 4!*6!/(2*2*2)*(6-1 choose 4) such anagrams (the 4! accounts for all the possible ways to order the 4 consonants and the 6!/(2*2*2) accounts for all the possible ways to order the 6 vowels (where we need to divide by 2 thrice because all three distinct vowels appear twice))
  • Only the first letter is a consonant. Using stars and bars again we find that there are 4!*6!/(2*2*2)*(6-1 choose 3) such anagrams
  • Only the last letter is a consonant. By symmetry there are 4!*6!/(2*2*2)*(6-1 choose 3) such anagrams as well
  • Both the first and the last letter are consonants. Using stars and bars one last time we find that there are 4!*6!/(2*2*2)*(6-1 choose 2) such anagrams
All in all, there are therefore 4!*6!/(2*2*2)*((5 choose 1)+3*(5 choose 2)) = 3*6!*(5+3*10) = 3*6!*35 = 15*7! = 15*5040 = 75600 anagrams without consecutive consonants
Giga mogger iq
 
y=|lnx|

Find the first order derivative of the above function
For x < 1, y = -ln x, so its derivative is -1/x. For x > 1, y = ln x, so its derivative is 1/x. For x = 1, the derivative does not exist.
 
For x < 1, y = -ln x, so its derivative is -1/x. For x > 1, y = ln x, so its derivative is 1/x. For x = 1, the derivative does not exist.


@trying to ascend help bro
 
For x < 1, y = -ln x, so its derivative is -1/x. For x > 1, y = ln x, so its derivative is 1/x. For x = 1, the derivative does not exist.
Find all functions R (greater than 0)---->R (greater than 0), such that:


f(xy + f(x)) = f(f(x)f(y)) + x
 


@trying to ascend help bro

As he pointed out, modulus is just the sqrt raised to the power of two (or any even root raised to its inverse),

So you just have to apply the chain rule and you will find the value, though you have to restrict the domain for it to work
 
As he pointed out, modulus is just the sqrt raised to the power of two (or the root of any even number raised to its inverse),

So you just have to apply the chain rule and you will find the value, though you have to restrict the domain for it to work
Ah okay
 
his answer is the same as mine, just way more complicated
Yeah but I've been using him and another YouTube channel called blackpenredpen to revise for my final math exam
 
Find all functions R (greater than 0)---->R (greater than 0), such that:


f(xy + f(x)) = f(f(x)f(y)) + x
You got a hint? I think I know the answer, but I ain't seeing how to prove it.
 
MATH SUCKS!!!!!!!!!!! I HAAAAAAAAAAAAAAAAATTTTTTTTTTTEEEEEEEEEEEE MAAAAAAAAAAAAAAAAAAAAAAAAAAATTTTTTTTTTTTTTTTHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 
All logs are in base 2.

Knowing that x belongs to the closed interval (1, 64), find the maximum value of the follwing function

F(x) = (logx) to the power of 4 + 12(logx) to the power of 2 . log(8/x)
 
Be Z a complex number. Given that the absolute value of Z is sqrt(3)/3 and that Z satisfies the following relation:

2(z - 1) to the power of 2017 = (sqrt(3) + i)(iZ - 1) to the power of 2017 (last only parenthesis in that case)

Find the value of z
 
All logs are in base 2.

Knowing that x belongs to the closed interval (1, 64), find the maximum value of the follwing function

F(x) = (logx) to the power of 4 + 12(logx) to the power of 2 . log(8/x)
Let u = lg x. Then F(x) = u^4 + 12u^2(3 - u) = u^2(u^2 - 12u + 36) = u^2(u - 6)^2 = (u(6 - u))^2, which is clearly maximal when u = 3, in which case the function takes the value 9^2 = 81.
 
Let u = lg x. Then F(x) = u^4 + 12u^2(3 - u) = u^2(u^2 - 12u + 36) = u^2(u - 6)^2 = (u(6 - u))^2, which is clearly maximal when u = 3, in which case the function takes the value 9^2 = 81.
Correct :feelsokman:
 
I need someone to check my answer

Find y prime:

x^y=y^x

Ans: y prime = [ln(y)-y/x]/[ln(x)-x/y]
 
Three players sit in a desk to play a game.

Alternately, each player rolls a non vicious cubic dice and, after each player rolls it, the player to its right gets to play.

If 1 is rolled, then the player to its left gets to play.

The one who rolls 6 wins the game.

Given that, what are the odds of the player that played first to win the game (not necessarily in his first play)?
 
Three players sit in a desk to play a game.

Alternately, each player rolls a non vicious cubic dice and, after each player rolls it, the player to its right gets to play.

If 1 is rolled, then the player to its left gets to play.

The one who rolls 6 wins the game.

Given that, what are the odds of the player that played first to win the game (not necessarily in his first play)?
using Markov chain methods I arrive at 32/79
 
Circumference C has equation x² + y² = 16. Let C' be a circle of radius 1 that moves internally tangent to circle C, without slipping between the contact points, that is, C' rolls internally on C.

4d603d17571c7a35dadd31d7348bf761



Point P on C' is defined so that at the beginning of the movement of C', point P coincides with the point of tangency (4,0), as shown in figure a. After a certain displacement, the angle between the x axis and the straight line joining the center of the circles is α, as shown in figure b.

Determine the coordinates of the point P marked on C' as a function of the angle α.
 
Circumference C has equation x² + y² = 16. Let C' be a circle of radius 1 that moves internally tangent to circle C, without slipping between the contact points, that is, C' rolls internally on C.

4d603d17571c7a35dadd31d7348bf761



Point P on C' is defined so that at the beginning of the movement of C', point P coincides with the point of tangency (4,0), as shown in figure a. After a certain displacement, the angle between the x axis and the straight line joining the center of the circles is α, as shown in figure b.

Determine the coordinates of the point P marked on C' as a function of the angle α.
:dafuckfeels: I think I'll return to studying for my physics paper
 
:dafuckfeels: I think I'll return to studying for my physics paper
Here is a physics problem:

Consider a capacitor octahedron, such that there is a capacitor between each vertex of capacitance C.

Each pair of vertex, sharing an edge or not, has a capacitor between it.

Calculate the equivalent capacitance between two neighbor vertex of the solid
 
Here is a physics problem:

Consider a capacitor octahedron, such that there is a capacitor between each vertex of capacitance C.

Each pair of vertex, sharing an edge or not, has a capacitor between it.

Calculate the equivalent capacitance between two neighbor vertex of the solid

:dafuckfeels:
 
Circumference C has equation x² + y² = 16. Let C' be a circle of radius 1 that moves internally tangent to circle C, without slipping between the contact points, that is, C' rolls internally on C.

4d603d17571c7a35dadd31d7348bf761



Point P on C' is defined so that at the beginning of the movement of C', point P coincides with the point of tangency (4,0), as shown in figure a. After a certain displacement, the angle between the x axis and the straight line joining the center of the circles is α, as shown in figure b.

Determine the coordinates of the point P marked on C' as a function of the angle α.
x = 3cos(α) + cos(-3α) = 3cos(α) + cos(3α) & y = 3sin(α) + sin(-3α) = 3sin(α) - sin(3α)
 
YOU WON'T FIND ANY OOGAA KARABOGAS HERE!!!!!!!!!!!!! FUCKING NIGGERS
 
@trying to ascend

hit me up with a math problem. i have an exam in a few minutes, but gimme something fun to come back to. i'll probably seriously keep up with this math problem thread just like the old one i used to do with @Divergent_Integral
 
@trying to ascend

hit me up with a math problem. i have an exam in a few minutes, but gimme something fun to come back to. i'll probably seriously keep up with this math problem thread just like the old one i used to do with @Divergent_Integral
Sure, though I'm studying more physics nowadays, that's why the thread is often empty of new problems.

Find the function f(x), such that

f((x-3)/(x + 1)) + f((x + 3)/(1 - x)) = x
 
Consider the ellipse below, where DD' is a chord passing through its center, MM' is a locus chord, and the major axis of the ellipse is 2a.

Prove that: DD' ²= MM' . 2a


2e256071e436487e776e40241e7cc2cf
 
Why would i care about this nerd shit
 
Why would i care about this nerd shit
Pass a hard entrance exam, become a STEM graduate in a good Uni, make >150k working at a big tech company
 
Pass a hard entrance exam, become a STEM graduate in a good Uni, make >150k working at a big tech company
@Ahnfeltia

oh whoops for some reason i thought this was the "Fun facts about incels" thread
 
@Ahnfeltia

oh whoops for some reason i thought this was the "Fun facts about incels" thread
You don't say. Your post was pretty apt tho. I just wanted to make a shitty vaguely-math-related quip.
 
I should be right at home here with a masters in math but I don't know how to solve the penultimate problem, and I was always weak in geometry. I'm really good at calculus, though. Taking a derivative or inverse just seems to make it worse.
 

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